When a battery sends current through a resistance $R_{1}$ for time ‘t’ , the heat produced in the resistor is Q. When the same battery sends current through another resistance $R_{2}$ for time ‘t’ , the heat produced in $R_{2}$ is again ‘Q’. The internal resistance of battery is

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$\frac{R_{1} {+R}_{2}}{2}$

$\sqrt{R_{1} R_{2}}$

$\frac{R_{1} {- R}_{2}}{2}$

$\frac{\sqrt{R_{1} {+R}_{2}}}{2}$

answer is C.

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Detailed Solution

${Q=i}_{1}^{2} R_{1} {t = i}_{2}^{2} R_{2} t \Rightarrow {(\frac{E}{{r + R}_{1}})}^{2} R_{1} = {(\frac{E}{{r + R}_{2}})}^{2} R_{2} ⇒ (\frac{E}{{r + R}_{1}}) \sqrt{R_{1}} = (\frac{E}{{r + R}_{2}}) \sqrt{R_{2}} \Rightarrow r \sqrt{R_{1}} + \sqrt{R_{1}} . R_{2} = r \sqrt{R_{2}} + \sqrt{R_{2}} . R_{1} \Rightarrow r \sqrt{R_{1}} - r \sqrt{R_{2}} = \sqrt{R_{2}} . R_{1} - \sqrt{R_{1}} . R_{2} \Rightarrow r (\sqrt{R_{1}} - \sqrt{R_{2}}) = \sqrt{R_{2}} \sqrt{{R_{1}}^{2}} - \sqrt{R_{1}} \sqrt{{R_{2}}^{2}} \Rightarrow r = \frac{\sqrt{R_{1} R_{2}} [\sqrt{R_{1}} - \sqrt{R_{2}}]}{[\sqrt{R_{1}} - \sqrt{R_{2}}]} \Rightarrow r = \sqrt{R_{1} R_{2}}$